x-intercepts: find x such that x^2-2x-3 = 0
y-intercept: plug in x=0 and find y
axis of symmetry: x = -b/2a
vertex: note that x^2-2x-3 = (x^2-2x+1)-4 = (x-2)^2 - 4
That should get you going.
y=x^2 - 2x - 3
y-intercept?
x-intercept?
axis of symmetry?
vertex?
5 answers
I don't have the brain capacity to get anything besides the y-intercept
If that's true, you are in deep doo doo
x^2-2x-3 = 0
(x-3)(x+1) = 0
x = -1, 3
-b/2a = 2/2 = 1
so the line x=1 is the axis of symmetry.
Note that it is midway between the x-intercepts
This is the vertex form of the equation.
Oops. It should be y = (x-1)^2 - 4
You can just read off the vertex.
It is at (1,-4)
Better get back to studying.
x^2-2x-3 = 0
(x-3)(x+1) = 0
x = -1, 3
-b/2a = 2/2 = 1
so the line x=1 is the axis of symmetry.
Note that it is midway between the x-intercepts
This is the vertex form of the equation.
Oops. It should be y = (x-1)^2 - 4
You can just read off the vertex.
It is at (1,-4)
Better get back to studying.
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Help y=-x^2+2x-5
y=−x
2
+2x−5
y=−x
2
+2x−5
1 answer